Distance between A and B is 230 km. After travelling a certain distance a motobike breaks down and then it travels at the 3/4th of the actual speed and reaches 1 hour late. If the bike had broken down after travelling a distance of 30 km more he would have reached 12 minutes earlier. Find out the distance at which the bike broke down and speed of bike.

Answer

To solve the problem, we will denote the following variables:

- Let \( D \) be the distance at which the bike broke down.
- Let \( S \) be the actual speed of the bike.
- The total distance from A to B is 230 km.

### Step 1: Set up the equations based on the problem statement.

1. The time taken to travel the distance \( D \) at speed \( S \) is \( \frac{D}{S} \).
2. After the breakdown, the remaining distance is \( 230 - D \), and the speed is \( \frac{3}{4}S \). The time taken for this part is \( \frac{230 - D}{\frac{3}{4}S} = \frac{4(230 - D)}{3S} \).
3. The total time taken when the bike breaks down at distance \( D \) is:
\[ T_1 = \frac{D}{S} + \frac{4(230 - D)}{3S} \]

### Step 2: Calculate the time taken if the bike had broken down 30 km earlier.

1. If the bike had broken down after traveling \( D + 30 \) km, the time taken for this distance is \( \frac{D + 30}{S} \).
2. The remaining distance is \( 230 - (D + 30) = 200 - D \), and the time taken for this part is \( \frac{200 - D}{\frac{3}{4}S} = \frac{4(200 - D)}{3S} \).
3. The total time taken in this case is:
\[ T_2 = \frac{D + 30}{S} + \frac{4(200 - D)}{3S} \]

### Step 3: Set up the equations based on the time differences.

1. According to the problem, the bike reaches 1 hour late when it breaks down at distance \( D \):
\[ T_1 = T_{actual} + 1 \]
2. If it had broken down 30 km earlier, it would have reached 12 minutes earlier (which is \( \frac{1}{5} \) hour):
\[ T_2 = T_{actual} - \frac{1}{5} \]

### Step 4: Equate the two expressions for time.

From the above equations, we can set up the following:

1. From \( T_1 = T_{actual} + 1 \):
\[ \frac{D}{S} + \frac{4(230 - D)}{3S} = T_{actual} + 1 \]
2. From \( T_2 = T_{actual} - \frac{1}{5} \):
\[ \frac{D + 30}{S} + \frac{4(200 - D)}{3S} = T_{actual} - \frac{1}{5} \]

### Step 5: Solve the equations.

1. Rearranging both equations gives us two expressions for \( T_{actual} \):
\[ T_{actual} = \frac{D}{S} + \frac{4(230 - D)}{3S} - 1 \]
\[ T_{actual} = \frac{D + 30}{S} + \frac{4(200 - D)}{3S} + \frac{1}{5} \]

2. Set the two expressions for \( T_{actual} \) equal to each other and simplify:
\[ \frac{D}{S} + \frac{4(230 - D)}{3S} - 1 = \frac{D + 30}{S} + \frac{4(200 - D)}{3S} + \frac{1}{5} \]

3. Cross-multiply and simplify to find \( D \) and \( S \).

### Step 6: Solve for \( D \) and \( S \).

After simplifying the equations, we will find the values of \( D \) (the distance at which the bike broke down) and \( S \) (the speed of the bike).